A capacitor has parallel plates of area 14 cm 2 separated by 1.6 mm. The space b
ID: 1317787 • Letter: A
Question
A capacitor has parallel plates of area 14 cm2 separated by 1.6 mm. The space between the plates is filled with polypropylene (see the table below).
(a) Find the permittivity of polypropylene.
C2/(N ? m2)
(b) Find the maximum permissible voltage across the capacitor to avoid dielectric breakdown.
V
(c) When the voltage equals the value found in part (b), find the surface-charge density on each plate and the induced surface-charge density on the surface of the dielectric.
Explanation / Answer
k = e/eo...
e = k*eo = 2.8*8.85*10^-12= 2.48*10^-11 C^2/(N-m^2)...
B) V = E*d = 3*10^7*1.6*10^-3 = 4.8*10^4 V
C) C = Q/V = (eo*A/d)*k = ((8.85*10^-12*14*10^-4)/(1.6*10^-3))*2.8 = 2.2*10^-11....
Q = 2.2*10^-11*48000 = 1.05*10^-6 C....
without dielectric = 1.05*10^-6*2.8 = 2.95*10^-6 C
surface charge density on each plate = (1.05*10^-6)/(14*10^-4) = 75*10^-5 C/m^2
on dielectric = 1.05*10^-6*(1.8-1)/(14*10^-4) = 6*10^-4 C/m^2
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